While working with a definite integral, I came across the following series:
$$g_n(x)=\sum\limits_{0\leqslant k<n/2} (-1)^k {n \choose 2k+1} x^{n-(2k+1)}\\ = \frac{i}{2}[(x-i)^n-(x+i)^n] \tag{1}$$
Here, the values of k are integers less than n/2. For the first three values of n, the values of the function are as follows:
$g_1(x)=1 \\g_2(x)=2x \\g_3(x)=3x^2-1$
From the second form of the series in equation (1), it can be readily obtained that
$$g_n(x)=\frac{i}{2}[(x-i)^n-(x+i)^n]=(1+x^2)^{n/2} \sin(x\sin^{-1}(\frac{1}{\sqrt{1+x^2}})) \tag{2}$$
I have two questions as follows:
- How is the second form of $g_n(x)$ obtained from the first form in equation (1)?
- Is it possible to obtain the result in equation (2) directly from the first form of equation (1), i.e. without using the expression $\frac{i}{2}[(x-i)^n-(x+i)^n]$?
I will assume that $n=2m$ is even. Since $a^2-b^2=(a+b)(a-b)$, note that $$(x-i)^n-(x+1)^n=((x-i)^m)^2-((x+i)^m)^2=(x-i)^m(x+i)^m=((x-i)(x+i))^m=(x^2+1)^m.$$ Now use that $$(a+b)^m=\sum\limits_{k=0}^m\binom{m}{k}a^nb^{m-k}$$ which yields for $b=x^2, a=1$ the formula $$(x^2+1)^m=\sum\limits_{k=0}^m\binom{m}{k}x^{2m-2k}.$$ Of course, since you are interested in $\frac{i}{2}((x-i)^n-(x+i)^n)$, you have to multiply the sum above by the factor $i/2$. Can you conclude?